Answer:
You need to put the diagrams. repost and include the diagrams
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
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hope this helped you :)
The answer to your question is false
First, figure out what the angular speed of the merry-go-around is.
v ϖ = Using the formula for linear speed Since the angular speed is constant, there is no angular acceleration. Tangential acceleration is t a Radial acceleration is r a Thus the total acceleration is a = rω =1.2m×1.6rad / s =1.9m/ s = rα =1.2m×0rad / s2 = 0m/ s2 =rϖ2 =1.2m×(1.6rad / s)2 =3.1m/ s2 2 2 t r = a +a = 0+(3.1)2 =3.1m/ s2 1 4.0 rev